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Math Help - Differentials:Please to if my work is correct

  1. #1
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    Differentials:Please to if my work is correct

    I just want to see if my work is correct.

    This problems comes in for parts. I am concerned about the last three. I think the processes should be longer but I don't see how I can make them longer.





    4.8 #1
    y`` + y = 0

    a.



    e^xsolve with: e^x
    e^x y`` + y = 0

    a y"+b y' +c y=0

    a y +b y + c =0

    a e^(y x) + b e^(y x)


    e^(y)(a cos(x) +b sin(x))

    This is not a solution


    b.Sin x
    sin x y'' + y = 0


    1/2 i e^-i^x y``(x)- 1/2 i e^i^x y``(x)+y(x)=0
    y(x)+sin(x) = cos(x) y''(x)
    Solution


    c.Cos x
    cos x y'' + y = 0

    1/2 e^(-i x) y''(x)+1/2 e^(i x) y''(x)+y(x) = 0cos(x) y''(x)+y(x) = 0Solution






    d.Sin x cos x

    -1/2 e^(-i x) y''(x)-1/2 e^(i x) y''(x)+y(x)+1/2 i e^(-i x)-1/2 i e^(i x) = 0


    Solution
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  2. #2
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    Quote Originally Posted by pikachumanson View Post
    I just want to see if my work is correct.

    This problems comes in for parts. I am concerned about the last three. I think the processes should be longer but I don't see how I can make them longer.





    4.8 #1
    y`` + y = 0

    a.



    e^xsolve with: e^x
    e^x y`` + y = 0

    a y"+b y' +c y=0

    a y +b y + c =0

    a e^(y x) + b e^(y x)

    e^(y)(a cos(x) +b sin(x))

    This is not a solution


    b.Sin x
    sin x y'' + y = 0


    1/2 i e^-i^x y``(x)- 1/2 i e^i^x y``(x)+y(x)=0
    y(x)+sin(x) = cos(x) y''(x)
    Solution


    c.Cos x
    cos x y'' + y = 0

    1/2 e^(-i x) y''(x)+1/2 e^(i x) y''(x)+y(x) = 0cos(x) y''(x)+y(x) = 0Solution






    d.Sin x cos x

    -1/2 e^(-i x) y''(x)-1/2 e^(i x) y''(x)+y(x)+1/2 i e^(-i x)-1/2 i e^(i x) = 0


    Solution
    Sorry but I can't make any sense of this. Please post the questions exactly as they are written from wherever you got them from.
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  3. #3
    MHF Contributor

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    [QUOTE=pikachumanson;533340]I just want to see if my work is correct.

    This problems comes in for parts. I am concerned about the last three. I think the processes should be longer but I don't see how I can make them longer.





    4.8 #1
    y`` + y = 0
    Is this the differential equation you want to solve?

    [quote[ a.



    e^xsolve with: e^x
    What does that mean?

    e^x y`` + y = 0
    Just put e^x in front of y''? Why?

    a y"+b y' +c y=0

    a y’’ +b y’ + c =0

    a e^(y’’ x) + b e^(y’ x)
    Now you have taken the exponential of each derivative. Again, why? And where did a, b, and c suddenly come from?

    e^(y)(a cos(x) +b sin(x))

    This is not a solution


    b.Sin x
    sin x y'' + y = 0


    1/2 i e^-i^x y``(x)- 1/2 i e^i^x y``(x)+y(x)=0
    y(x)+sin(x) = cos(x) y''(x)
    Solution


    c.Cos x
    cos x y'' + y = 0

    1/2 e^(-i x) y''(x)+1/2 e^(i x) y''(x)+y(x) = 0cos(x) y''(x)+y(x) = 0Solution






    d.Sin x – cos x

    -1/2 e^(-i x) y''(x)-1/2 e^(i x) y''(x)+y(x)+1/2 i e^(-i x)-1/2 i e^(i x) = 0


    Solution
    I agree with mr. fantastic- what you wrote makes no sense at all!
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